Solution :
Demand for cola : 100 – 34x + 5y
Demand for cola : 50 + 3x – 16y
Therefore, total revenue :
x(100 – 34x + 5y) + y(50 + 3x – 16y)
R(x,y) = ![$100x-34x^2+5xy+50y+3xy-16y^2$](https://tex.z-dn.net/?f=%24100x-34x%5E2%2B5xy%2B50y%2B3xy-16y%5E2%24)
![$R(x,y) = 100x-34x^2+8xy+50y-16y^2$](https://tex.z-dn.net/?f=%24R%28x%2Cy%29%20%3D%20100x-34x%5E2%2B8xy%2B50y-16y%5E2%24)
In order to maximize the revenue, set
![$R_x=0, \ \ \ R_y=0$](https://tex.z-dn.net/?f=%24R_x%3D0%2C%20%5C%20%5C%20%5C%20R_y%3D0%24)
![$R_x=\frac{dR }{dx} = 100-68x+8y$](https://tex.z-dn.net/?f=%24R_x%3D%5Cfrac%7BdR%20%7D%7Bdx%7D%20%3D%20100-68x%2B8y%24)
![$R_x=0$](https://tex.z-dn.net/?f=%24R_x%3D0%24)
.............(i)
![$R_y=\frac{dR }{dx} = 50-32x+8y$](https://tex.z-dn.net/?f=%24R_y%3D%5Cfrac%7BdR%20%7D%7Bdx%7D%20%3D%2050-32x%2B8y%24)
![$R_y=0$](https://tex.z-dn.net/?f=%24R_y%3D0%24)
.............(ii)
Solving (i) and (ii),
4 x (i) ⇒ 272x - 32y = 400
(ii) ⇒ (-<u>) 8x - 32y = -50 </u>
264x = 450
∴ ![$x=\frac{450}{264}=\frac{75}{44}$](https://tex.z-dn.net/?f=%24x%3D%5Cfrac%7B450%7D%7B264%7D%3D%5Cfrac%7B75%7D%7B44%7D%24)
![$y=\frac{175}{88}$](https://tex.z-dn.net/?f=%24y%3D%5Cfrac%7B175%7D%7B88%7D%24)
So, x ≈ $ 1.70 and y = $ 1.99
R(1.70, 1.99) = $ 134.94
Thus, 1.70 dollars per cola
1.99 dollars per iced ted to maximize the revenue.
Maximum revenue = $ 134.94
Answer:
40000 and 12.5% respectively
Step-by-step explanation:
soln:
given,
cost price(c.p)=320000
selling price(s.p)=280000
loss(l),loss percentage(l%)=?
we know,
l=c.p-s.p
or,l=320000-280000
•°•l=40000
now,
l%=l/c.p*100%
or,l%=40000/320000*100%
•°•l%=12.5%